Exact Thin-walled Curved Beam Element Considering Shear Deformation Effects and …C).pdf · 2018....

Steel Structures 6 (2006) 191-207 www.kssc.or.kr

Exact Thin-walled Curved Beam Element Considering Shear

Deformation Effects and Non-symmetric Cross Sections

Nam-Il Kim1, Chung C. Fu1 and Moon-Young Kim2,*

1Faculty Research Associate, Department of Civil and Environmental Engineering, University of Maryland,

College Park, MD 20742, USA2Department of Civil and Environmental Engineering, Sungkyunkwan University,

Cheoncheon-Dong, Jangan-Ku, Suwon, 440-746, S. Korea

Abstract

A thin-walled and spatially coupled exact curved beam element subjected to initial axial force is developed considering theshear deformation effects and the non-symmetric cross sections. For this purpose, a shear deformable thin-walled curved beamtheory is presented considering the warping effect. Next, equilibrium equations for a curved beam are transformed into firstorder simultaneous ordinary differential equations by introducing 14 displacement parameters. Then the exact displacementfunction is obtained by determining homogeneous solutions of simultaneous differential equations via generalized eigenproblemwith non-symmetric matrix. Lastly, the exact static element stiffness matrix is evaluated using force-deformation relationships.In order to demonstrate the accuracy and practical usefulness of this method, the displacements and the normal stresses of thin-walled curved beams are evaluated and compared with those by thin-walled curved beam elements as well as shell elements.

Keywords: curved beam, non-symmetric cross section, exact stiffness matrix, warping, shear deformation

1. Introduction

For the static analysis of curved structures, improved

curved beam theories have generated a lot of interest

among researchers in recent years. Modeling of curved

structures by means of lower-order isoparametric beam

elements leads to excessively stiff behavior (called shear

locking) in the thin regimes. Classical curved beam

elements, when used for modeling thin and deep arches

also exhibited excessive bending stiffness (called membrane

locking) in approximating inextensional bending response.

Up to the present, a large amount of work was devoted to

the improvement of curved beam elements in order to

overcome these shear and membrane locking phenomena

and to obtain acceptable results for coarse meshes.

Reduced integration (Prathap 1985, Stolarski and Belytschko

1983, 1982, Prathap and Bhashyam 1982) of shear and

membrane energies is widely used for eliminating one or

more higher-order components in the strain distribution

which leads to spurious kinematic modes in their respective

thin limits. However, indiscriminate use of reduced

integration can introduce zero energy modes. Babu and

Prathap (1986), Prathap and Babu (1986) proposed a

field-consistency approach, which identifies the spurious

constraints of the inconsistent strain field and drops them

in advance. Unlike reduced integration method, a field

consistency approach ensures a variationally correct and

orthogonally consistent strain field. But both these

methods reduce the order of strain interpolation and suffer

from lower convergence rate. Curved beam elements based

on displacement fields derived from assumed independent

strain fields exhibited no locking behavior (Lee and Sin

1994, Choi and Lim 1995, 1993). Applying the assumed

polynomials for the strain fields, the strain-displacement

relations are solved to get general solutions for the

displacement fields.

On the other hand, a few researchers have been

interested in development of curved beam element using

the displacement fields which satisfy a homogeneous

form of equilibrium equations of a curved beam element.

Raveendranath et al. (1999) developed two-noded locking-

free curved beam elements, for which a cubic polynomial

field was assumed a priori and the polynomial interpolations

for the axial displacement and the twisting angle were

derived employing force-moment and moment-shear

equilibrium equations. Recently Zhang and Di (2003)

presented the new, accurate two-noded finite elements

which are free from shear and membrane locking and are

derived from the potential energy principle and the

Hellinger-Reissner functional principle, respectively.

However, most of these studies are restricted to two

dimensional problems and based on explicitly analytical

*Corresponding authorTel: +82-31-290-7514, Fax: +82-31-290-7548E-mail: [email protected]

192 Nam-Il Kim et al.

solutions of homogeneous equations.

Even though a significant amount of research has been

conducted on development of an improved curved beam

element for the static analysis, to the authors’ knowledge

there was no study evaluating the exact static element

stiffness matrix of spatially coupled shear deformable

thin-walled curved beams with non-symmetric cross

section via generalized eigenvalue problem. It is well

known that the elastic behavior of thin-walled curved

beams with non-symmetric cross section is very complex

due to the coupling effect of extensional, bending, and

torsional deformation and many researchers thought that

it is often difficult and sometimes impossible to solve the

spatially coupled curved beam problem exactly due to an

aforementioned reason.

The primary aim of this study is to present an effective

method of evaluating the exact static element stiffness

matrix based on a shear deformable and non-symmetric

thin-walled curved beam theory with initial axial force

including the thickness-curvature effect and warping

deformation. For this purpose, equilibrium equations and

force-deformation relations are first derived for a uniform

curved beam element. Next, higher order differential

equations are transformed into a set of the first-order

simultaneous ordinary differential equations by introducing

14 displacement parameters. Then, a generalized linear

eigenproblem with complex eigenvalues is established

and explicit expressions for displacement parameters are

calculated by solving the simultaneous homogeneous

ordinary differential equations. Finally, the exact static

element stiffness matrix is evaluated using force-deformation

relations. In addition, an isoparametric curved beam element

having two nodes is presented based on the same total

potential energy as an exact curved beam element.

In order to demonstrate the accuracy and the practical

usefulness of this study, the displacements and the normal

stresses at the arbitrary points of non-symmetric cross

section for the spatially coupled curved beam are evaluated

and compared with FE solutions using curved beam

elements and shell elements.

2. Shear Deformable Curved Beam Theory Considering Non-symmetric Thin-walled Cross Sections

Force-deformation relationships and equilibrium equations

of a shear deformable curved beam having non-symmetric

thin-walled cross sections are derived in this Section.

2.1. Force-deformation relationships

To derive the equilibrium equations of a shear deformable

curved beam with non-symmetric cross section, a global

curvilinear coordinate system (x1, x2, x3), as shown in Fig.

1, is adopted in which the x1 axis coincides with a

centroid axis having the radius of curvature R but x2, x3are not necessarily principal inertia axes. Figs. 2(a) and

2(b) show the displacement parameters and the stress

resultants of thin-walled curved beams defined at the

non-symmetric cross-section, respectively. Ux, Uy, Uz and

ω1, ω2, ω3 are the rigid body translations and the rotations

of the cross section with respect to x1, x2 and x3 axes,

respectively. f is the displacement parameter measuring

warping deformations. , mean principal axes defined

at the centroid where α is the angle between and

x2

px3

p

x2

px2

Figure 1. Coordinate system of a thin-walled curved beam.

Figure 2. Notation for displacement parameters and stress resultants.

Exact Thin-walled Curved Beam Element Considering Shear Deformation Effects and Non-symmetric Cross Sections 193

axes in the counterclockwise direction. Assuming that the

cross section is rigid in its own plane, the total

displacement field can be written as follows

(1a)

(1b)

(1c)

where φ = the normalized warping function defined at

the centroid. And stress resultants with respect to the

centroid are defined as follows

(2a-i)

where F1, F2 and F3 = the axial and shear forces acting

at the centroid; M1= the total twist moment with respect

to the centroid axis; M2 and M3= the bending moments

with respect to x2 and x3 axes, respectively; MR and Mφ =

the restrained torsional moment and the bimoment about

the x1 axis, respectively; Mp = the stress resultant known

as the Wagner effect. Sectional properties are defined by

(3a-o)

where I2, I3, I23 and Iφ = the second moment of inertia

about x2 and x3 axes, the product moment of inertia and

the warping moment of inertia, respectively; Iφ2 (=I2e2)

and Iφ3 (=−I2e2) = the product moments of inertia due to

the normalized warping; Iijk (i, j, k = f, 2, 3) = the third

moments of inertia. Also, normal and shear strain-

displacement relations may be expressed as follows

(4a)

(4b)

(4c)

Now substituting Eq. (4a) into Eqs. (2a), (2e), (2f), (2g) and integrating over the cross section leads to

U1

Ux x2ω3– x3ω2 fφ x2 x3,( )+ +=

U2

Uy x3ω1–=

U3

Uz x2ω1–=

F1

τ11

AdA∫= F2 τ12 AdA∫= F3 τ13 AdA∫= M1 τ13x2 τ12x3–( ) AdA∫=,, ,

M2

τ11x3Ad

A∫= M3 τ11x2 A Mφ τ11φ AdA∫= Mp τ11 x22x3

2+( ) Ad

A∫=,,dA∫–=,

MR τ12φ 2, τ13 φ 3,φ

R x3

+------------–⎝ ⎠

⎛ ⎞+

R x3

+

R------------ Ad

A∫=

I2

x3

2Ad

A∫= I3 x22Ad

A∫= I23 x2x3 AdA∫= Iφ φ2A Iφ2 φx3 AdA∫=,dA∫=,, ,

Iφ3 φx2 AdA∫= I222 x33A I

223x2x3

2Ad

A∫= I233 x23x3Ad

A∫=,,dA∫–=,

I333

x2

3Ad

A∫= Iφ22 φx32A Iφ23 φx2x3 AdA∫= Iφ33 φx2

2Ad

A∫=,,dA∫–=,

Iφφ2 φ2x3Ad Iφφ3 φ

2x2Ad

A∫=,A∫=

e11

U1 1,

U3

R------+⎝ ⎠

⎛ ⎞ RR x

3+

------------ U'xUz

R------+⎝ ⎠

⎛ ⎞ x2ω'

3

ω1

R------–⎝ ⎠

⎛ ⎞– x

3ω'

2φf'+ +

R

R x3

+------------= =

2e12

U2 1, R

R x3

+------------- U

1 2,+=

U'y ω3–( )R

R x3

+------------ φ

2,

x3R

R x3

+------------+⎝ ⎠

⎛ ⎞ ω'1

ω3

R------+⎝ ⎠

⎛ ⎞ f ω'1

ω3

R------+ +⎝ ⎠

⎛ ⎞φ2,+–=

2e13

U3 1,

U1

R------–⎝ ⎠

⎛ ⎞ RR x

3+

------------ U1 3,+=

U'z ω2Ux

R------–+⎝ ⎠

⎛ ⎞ RR x

3+

------------x2R

R x3

+------------ φ

3,–φ

R x3

+------------+⎝ ⎠

⎛ ⎞ ω'1

ω3

R------+⎝ ⎠

⎛ ⎞ f ω'1

ω3

R------+ +⎝ ⎠

⎛ ⎞ φ3,

φR x

3+

------------–⎝ ⎠⎛ ⎞

+ +=


(5a-d)

where E = the Young’s modulus and

(6a-f)

The following approximation is used.

(7)

On the other hand, force-deformation relations due to

shear stresses are given by Kim (2003) as

(8a-c)

(8d)

where G = the shear modulus; J = the torsional constant

and

(9a-f)

where , and = the effective shear areas defined by

(10a-c)

and

(11a-f)

where φs = the normalized warping function defined at

the shear center.

2.2. Equilibrium equations

In case of the thin-walled curved beam, the total

potential energy Π is expressed as follows:

Π = ΠE + ΠG − Πext (12)

where ΠE, ΠG and Πext = the elastic strain energy, the

potential energies due to initial axial forces and the nodal

forces, respectively. Their detailed expressions are

(13a)

(13b)

where Ue, Fe = the nodal displacement and nodal force

vectors which are defined by Eq. (31) and Eq. (38),

respectively.

Substituting the linear strain Eq. (4) into Eq. (13a) and

integrating over the cross sectional area, Eq. (13a) is

reduced to the following equation.

(14)

Now substitution of the force-deformation relations (5) and (8) into Eq. (14) leads to the elastic strain energy

F1

M2

M3

Mφ⎩ ⎭⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

E

AÎ2

R2

-----+Î2

R----–

Î23

R------

Iˆφ2

R------–

Î2

R----– Iˆ2 I

ˆ23– I

ˆφ2

Î23

R------ Iˆ23– I

ˆ3 I

ˆφ2–

Îφ2

R------– Iˆφ2 I

ˆφ3– I

ˆφ

U'xUz

R------+

ω'2

ω'3

ω1

R------–

f'⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫

=

Iˆ2 I2I222

R--------–= Iˆ3 I3

I233

R--------–= Iˆ23 I23

I223

R--------–=, ,

Iˆφ IφIφφ2

R--------–= Iˆφ2 Iφ2

Iφ22

R--------–= Iˆφ3 Iφ3

Iφ23

R--------–=, ,

R

R x3

+------------ 1

x3

R----–

x3

R----⎝ ⎠⎛ ⎞

2

+≅

F2

F3

MR⎩ ⎭⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

G

A2A23A2r

A23A3A3r

A2rA3r Ar

U'y ω3–

U'zUx

R------– ω

2+

ω'1

ω3

R------ f+ +

⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫

=

MST M1= Mr– M1= M'φ– GJ ω'1ω

3

R------+⎝ ⎠

⎛ ⎞=

A2

A2

scos

2α A2

ssin

2α+=

Ar ArsA2

se3

2A3

se2

2+ +=

A2r A2

se3cosα– A

3

se2sinα–=

,

,

,

A3

A3

scos

2α A2

ssin

2α+=

A23

A2

sA3

s+( )cosα sinα=

A3r A2

se3sinα– A

3

se2cosα–=

A2

sA3

sArs

1

A2

s-----

1

I3p

2------ Q

3

2 sd

t-----

A∫=1

A3

s-----

1

I2p

2------ Q

2

2 sd

t-----

A∫=1

Ars

-----1

Iφ2( )

2---------- Qr

2 sd

t-----

A∫=,,

I2p x3

p( )2

AdA∫= I3p x2

p( )2

AdA∫= Iφ

s φs( )2

AdA∫=,,

Q2

x3

pt sd

o

s

∫= Q3 x2pt sd

o

s

∫= Qr φst sd

o

s

∫=,,

ΠE1

2--- τ

11e11

2τ12e12

2τ13e13

+ +[ ]R x

3+

R------------ Ad x

1d

A∫ol

∫=

Πext1

2---U

e

TFe

=

ΠE1

2--- F

1[

o

l

∫ U'xUz

R------+⎝ ⎠

⎛ ⎞ M2ω'

2M

3ω'

3

ω1

R------–⎝ ⎠

⎛ ⎞– Mφf' F2+ + U'y ω3–( )+=

F3U'z

Ux

R------– ω

2+⎝ ⎠

⎛ ⎞ M1MR–( ) ω'1

ω3

R------–⎝ ⎠

⎛ ⎞ MR ω'1ω

3

R------ f+ +⎝ ⎠

⎛ ⎞ dx1

+ + +


(15)

The potential energy due to the initial axial stress can

be expressed as follows

(16)

where

(17)

Now, by variation of Eq. (12) with respect to seven

displacements, the equilibrium equations and the

boundary conditions for shear deformable curved beam

are derived as follows

(18a)

(18b)

(18c)

(18d)

(18e)

(18f)

(18g)

and

ΠE1

2--- EA U'x

Uz

R------+⎝ ⎠

⎛ ⎞2

EÎ2 ω'2U'x

R-------

Uz

R2

------––⎝ ⎠⎛ ⎞ EÎ3 ω'3

ω1

R------–⎝ ⎠

⎛ ⎞2

EÎφf'2

2EÎφ2 ω'2U'x

R-------

Uz

R2

------––⎝ ⎠⎛ ⎞f' 2EÎφ3 ω'3

ω1

R------–⎝ ⎠

⎛ ⎞f'–+ + ++o

l

∫=

2EÎ23 ω'3ω

1

R------–⎝ ⎠

⎛ ⎞ ω'2

U'x

R-------

Uz

R2

------––⎝ ⎠⎛ ⎞ GJ ω'

1

ω3

R------–⎝ ⎠

⎛ ⎞2

GA2U'y ω3–( )

2

GA3U'z

Ux

R------– ω

2+⎝ ⎠

⎛ ⎞2

+ + + GAr ω'1ω

3

R------ f+ +⎝ ⎠

⎛ ⎞2

+–

2GA23U'y ω3–( ) U'z

Ux

R------– ω

2+⎝ ⎠

⎛ ⎞2GA

2r U'y ω3–( ) ω'1ω

3

R------ f+ +⎝ ⎠

⎛ ⎞2GA

3r U'zUx

R------– ω

2+⎝ ⎠

⎛ ⎞ ω'1

ω3

R------ f+ +⎝ ⎠

⎛ ⎞ dx1

+ + +

ΠG1

2--- F

o

1U'y

2

U'zUx

R------+⎝ ⎠

⎛ ⎞2

β ω'1

ω3

R------–⎝ ⎠

⎛ ⎞2

+ + dx1o

l

∫=

βÎ2 Î3+

AÎ2

R2

-----+

-------------=

EA U''x1

R---U'z+⎝ ⎠

⎛ ⎞ 1R---EÎ2 ω''2

1

R---U''x–

1

R2

-----U'z–⎝ ⎠⎛ ⎞ 1

R---EÎφ2f''

1

R---EÎ23 ω''3

1

R---ω'

1–⎝ ⎠

⎛ ⎞ 1R---GA

3U'z ω2–

1

R---Ux+⎝ ⎠

⎛ ⎞––+ +–

1

R---GA

23U'y ω3–( )

1

R---GA

3r ω'1 f1

R---ω

3+ +⎝ ⎠

⎛ ⎞– F

o 1

R---U'z

1

R2

-----Ux–⎝ ⎠⎛ ⎞

0=––

GA2U''y ω'3–( ) GA23 U''z ω'2

1

R---U'x–+⎝ ⎠

⎛ ⎞ GA2r ω''1 f'

1

R---ω'

3+ +⎝ ⎠

⎛ ⎞ Fo1U''y+ + + 0=

1

R---EA U'x

1

R---Uz+⎝ ⎠

⎛ ⎞ 1

R2

-----– EÎ2 ω'21

R---U'x–

1

R2

-----Uz–⎝ ⎠⎛ ⎞ 1

R2

-----– EÎφ2f'1

R2

-----EÎ23 ω'31

R---ω

1–⎝ ⎠

⎛ ⎞ GA3U''z ω'2–

1

R---U'x+⎝ ⎠

⎛ ⎞–+

GA23U''y ω'3–( ) GA3r ω''1 f'

1

R---ω'

3+ +⎝ ⎠

⎛ ⎞– F

oU''z

1

R---U'x–⎝ ⎠

⎛ ⎞0=––

1

R---EI

ˆ3 ω'3

1

R---ω

1–⎝ ⎠

⎛ ⎞ 1R---EI

ˆφ3f'

1

R---EÎ23 ω'2

1

R---U'x–

1

R2

-----Uz–⎝ ⎠⎛ ⎞ GJ ω''

1

1

R---ω'

3+⎝ ⎠

⎛ ⎞–+ +–

GAr ω''1 f'1

R---ω'

3+ +⎝ ⎠

⎛ ⎞ GA2r U''y ω'3–( )– GA3r U''z ω'2

1

R---U'x–+⎝ ⎠

⎛ ⎞– β Fo

1ω''

1

1

R---ω'

3+⎝ ⎠

⎛ ⎞0=––

EÎ2 ω''21

R---U''x–

1

R2

-----U'z–⎝ ⎠⎛ ⎞ EI

ˆφ2f' EI

ˆ23 ω''3

1

R---ω'

1–⎝ ⎠

⎛ ⎞ GA3U'z ω2–

1

R---Ux+⎝ ⎠

⎛ ⎞ GA23U'y ω3–( ) GA3r ω'1 f

1

R---ω

3+ +⎝ ⎠

⎛ ⎞0=+ + + +––

EIˆ3 ω''3

1

R---ω'

1–⎝ ⎠

⎛ ⎞ EIˆφ3f'' EÎ23 ω''2

1

R---U''x–

1

R2

-----U'z–⎝ ⎠⎛ ⎞ 1

R---GJ ω''

1

1

R---ω'

3+⎝ ⎠

⎛ ⎞– GA

2U'y ω3–( ) GA23 U'z ω2–

1

R---Ux+⎝ ⎠

⎛ ⎞––+ +–

1

R---GAr+ ω'1 f

1

R2

-----ω3

+ +⎝ ⎠⎛ ⎞ GA

2r ω'1 f1

R---U'y–

2

R---ω

3+ +⎝ ⎠

⎛ ⎞–

1

R---GA

23U'z ω2

1

R---Ux–+⎝ ⎠

⎛ ⎞– β Fo

1

1

R---ω'

1

1

R2

-----ω3

+⎝ ⎠⎛ ⎞

0=–

EIˆφ2f'' EÎφ2 ω''2

1

R---U''x–

1

R2

-----U'z–⎝ ⎠⎛ ⎞ EÎφ3 ω''3

1

R---ω'

1–⎝ ⎠

⎛ ⎞ GAr ω'1 f1

R---ω

3+ +⎝ ⎠

⎛ ⎞ GA2r U'y ω3–( ) GA3r U'z ω2

1

R---Ux–+⎝ ⎠

⎛ ⎞0=+ + + +––


δUx(0) = or F1(0) = ; δUx(l) = or F1(l) = (19a,b)

δUy(0) = or F2(0) = ; δUy(l) = or F2(l) = (19c,d)

δUz(0) = or F3(0) = ; δUz(l) = or F3(l) = (19e,f)

δω1(0) = or M1(0) = ; δω1(l) = or M1(l) = (19g,h)

δω2(0) = or M2(0) = ; δω2(l) = or M2(l) = (19i,j)

δω3(0) = or M3(0) = ; δω3(l) = or M3(l) = (19k,l)

δf(0) = or Mφ(0) = ; δf(l) = or Mφ(l) = (19m,n)

Force-deformation relations are

(20a)

(20b)

(20c)

(20d)

(20e)

(20f)

(20g)

3. Exact Shear Deformable Curved Beam Element Having Non-symmetric Thin-walled Cross Sections

In this Section, an exact displacement function of thin-

walled curved beam is evaluated and its static element

stiffness matrix is calculated.

3.1. Exact evaluation of displacement functions

In order to transform the equilibrium equations in Eq.

(18) into a set of the first order ordinary differential

equations, a displacement state vector composed of 14

displacement parameters is defined by

d(x) = T

d(x) =


d'3 = d4 (22c)

(22d)

(22e)

(22f)

(22g)

(22h)

(22i)

(22j)

(22k)

(22l)

(22m)

(22n)

which can be compactly expressed as Eq. (23).

Ad' = Bd (23)

where components of matrices A and B are given in

Appendix I.

In order to find the homogeneous solution of the

simultaneous differential equation (23), the following

eigenvalue problem with non-symmetric matrix is taken

into account.

λAZ = BZ (24)

In practice, the general eigenvalue problem Eq. (24)

has the complex eigenvalue and the associated eigenvector

because the matrix A is symmetric but B is non-

symmetric. IMSL subroutine DGVCRG (Microsoft

IMSL Library 1995) is used to obtain the complex

eigensolutions of Eq. (24). From Eq. (24), 14 eigenvalues

λi and 14×14 eigenvectors Zi in complex domain can be

calculated.

(λi, Zi), i = 1, 2, ..., 14 (25)

GA2

Fo

1+( )d'

4GA

23d'

8GA

2rd'

12+ +

GA23

R------------d

2GA

2

GA3r

R-----------–⎝ ⎠

⎛ ⎞d6GA

23d10

– GA2rd14

–+=

d'5

d6

=

1

R--- EA

EÎ2

R2

------- GA3

Fo

1+ + +⎝ ⎠

⎛ ⎞d'

1GA

23d'

4–

EÎ23

R2

---------d'5

GA3

Fo

1+( )d'

8–

EÎ2

R2

-------d'9

– GA3rd'

12–+

GA23

GA3r

R-----------–⎝ ⎠

⎛ ⎞d6

1

R2

----- EAEÎ2

R2

-------+⎝ ⎠⎛ ⎞

d7

– GA3d10

EÎ23

R3

---------d11

GA3r

EÎφ2

R2

----------+⎝ ⎠⎛ ⎞

d14

+ + +–=

d'7

d8

=

1

R---EÎ23

R--------- GA

3r–⎝ ⎠

⎛ ⎞d'

1GA

2rd'

4–

1

R--- EÎ3 GAr+( )d'5– GA3rd'8–

EÎ23

R---------d'

9GJ GA

rβ Fo

1+ +( )d'

12–+–

GJ

R------- GA

2r–

βR--- F

o

1+⎝ ⎠

⎛ ⎞d6

EÎ23

R3

---------d7GA

3rd10

EÎ3

R2

-------d11

–EÎφ3

R---------- GA

r–⎝ ⎠

⎛ ⎞d14

–+ +=

d'9

d10

=

EÎ2

R-------d'

2GA

23d'

3EÎ23d'6 GA3d'7 EÎ2– d'10 GA3rd'11 EÎφ2– d'14+ + + +

GA3

R----------= d

1GA

23

GA3r

R-----------–⎝ ⎠

⎛ ⎞d5

EÎ2

R2

-------– d8GA

3– d

9

EÎ2

R-------d

12GA

3r– d

13+ +

d'11

d12

=

EÎ23

R---------d'

2GA

2

GA2r

R-----------–⎝ ⎠

⎛ ⎞d'

3– EÎ3d'6– GA23

GA3r

R-----------–⎝ ⎠

⎛ ⎞d'

7– EÎ23d'10

GJ

R-------

GAr

R--------- GA

2r–

βR--- F

o

1+ +⎝ ⎠

⎛ ⎞d'

12EIˆφ3d'14+ + +–

GA23

R------------

GA3r

R2

-----------–⎝ ⎠⎛ ⎞

d2

GJ

R2

------- GA2

GAr

R2

---------2GA

2r

R---------------–

β

R2

----- Fo

1+ + +⎝ ⎠

⎛ ⎞d5

–EÎ23

R2

---------d8

GA23

GA3r

R-----------–⎝ ⎠

⎛ ⎞d9

EÎ3

R-------d

12–

GAr

R--------- GA

2r–⎝ ⎠

⎛ ⎞d13

–+ +–=

d'13

d14

=

EÎφ2

R----------d'

2GA

2rd'

3EÎφ3d'6 GA3rd'7 EÎφ2– d'10 GArd'11 EI

ˆφd'14+ + + + +

GA3r

R-----------d

1GA

2r

GAr

R---------–⎝ ⎠

⎛ ⎞d5

EÎφ2

R2

----------– d8GA

3r– d

9

EÎφ3

R----------d

12GA

rd13

–+ +=


where

Zi = T (26)

Based on the above eigensolutions, it is possible that

the general solution of Eq. (23) is represented by the

linear combination of eigenvectors with complex exponential

functions as follows

(27)

where

a = T (28a)

(28b)

The a is the integration constant vector and X(x)

denotes the 14×14 matrix function made up of 14

eigensolutions.

Now, it is necessary that complex coefficient vector a

is represented with respect to 14 nodal displacement

components as shown in Fig. 3. For this, the nodal

displacement vector is defined by

Ue = T (29a)

Uα = < , , , , , , >T , α = p, q

(29b)

where

Up = T

(30a)

Uq = T (30b)

By substituting the coordinates of the member end

(x = 0, l) into Eq. (27) and accounting for Eq. (29), the

nodal displacement vector Ue is obtained as follows

Ue = Ea (31)

Finally, elimination of the complex coefficient vector a

from Eq. (31) and Eq. (27) yields the displacement state

vector.

d(x) = X(x)E−1Ue (32)

where X(x)E−1 denotes the exact interpolation matrix.

3.2. Calculation of static element stiffness matrix

Force-deformation relations in Eq. (20) of thin-walled

curved beam can be rewritten with respect to 14

displacement parameters in Eq. (21) as follows

(33a)

(33b)

(33c)

(33d)

(33e)

(33f)

(33g)

which is compactly represented as matrix form

f(x) = Sd(x) (34)

where f = T and each

component of 7 × 14 matrix S is presented in Appendix I.

Now, substituting Eq. (32) into Eq. (34) leads to

d x( ) aiZieλ

ix

i 1=

14

∑ X x( )a= =

X x( ) Z1eλ1x

Z2eλ2x

Z3eλ3x

; Z4eλ4x

; Z5eλ5x

; Z6eλ6x

; Z7eλ7x

; Z;8eλ8x

Z9eλ9x

; Z;10eλ10x

Z;11eλ11x

Z;12eλ12x

Z;13eλ13x

Z;14eλ14x

;[ ]=

Uxα

Uyα

Uzα ω

1

α ω2

α ω3

αfα

F1

EAEÎ2

R2

-------+⎝ ⎠⎛ ⎞d

2

EÎ23

R---------d

6

EA

R-------

EÎ2

R3

-------+⎝ ⎠⎛ ⎞d

7

EÎ2

R-------d

10–

EÎ23

R2

---------d11

–EÎφ2

R----------d

14–+ +=

F2

GA23

R------------d

1GA

2F

o

1+( )d

4GA

2

GA2r

R-----------–⎝ ⎠

⎛ ⎞d5

– GA23d8GA

23d9GA

2rd12 GA2rd13+ + + + +–=

F3

1

R--- GA

3F

o

1+( )d

1– GA

23d4

GA23

GA3r

R-----------–⎝ ⎠

⎛ ⎞d5

– GA3

Fo

1+( )d

8GA

3d9GA

3rd12 GA3rd13+ + + + +=

M1

GA3r

R-----------d

1GA

2rd4GJ

R-------

GAr

R--------- G–+ A

2r

βR--- F

o

1+⎝ ⎠

⎛ ⎞d5GA

3rd8 GA3rd9 GJ GAr β Fo

1+ +( )d

12GArd13+ + + + + +–=

M2

EÎ2

R-------d

2EÎ23– d6

EÎ2

R2

-------– d7EÎ2d10

EÎ23

R---------d

11EÎφ2d14+ + +–=

M3

EÎ23

R---------d

2EÎ3d6

EÎ23

R2

---------d7EÎ23d10–

EÎ3

R-------d

11– EÎφ3d14–+ +=

MφEÎφ2

R----------d

2– EÎφ3d6–

EÎφ2

R2

----------d7

– EÎφ2d10EÎφ3

R2

----------d11

EÎφd14+ + +=


f(x) = SX(x)E−1 Ue (35)

Also the nodal force vector as shown in Fig. 4 is

defined by

Fe = T (36a)

Fα = < , , , , , , >T , α = p, q

(36b)

Therefore, nodal forces at ends of element (x = 0, l) are

evaluated using Eq. (36) as

Fp = −f(0) = −SX(0)E−1 Ue (37a)

Fq = f(l) = SX(l)E−1 Ue (37b)

Consequently, the exact static stiffness matrix K of a

spatially coupled thin-walled curved beam element with

non-symmetric cross section subjected to initial axial

forces is evaluated as follows:

Fe = KUe (38)

where

(39)

3.3. Exact evaluation of normal stress of shear

deformable curved beam

To evaluate the normal stress at an arbitrary point of the

non-symmetric cross section of curved beam subjected to

external forces, the normal strain in Eq. (4a) can be

rewritten using the displacement parameters in Eq. (21)

as

(40)

Finally, the exact normal stress at an arbitrary point (x2,

x3) of the non-symmetric thin-walled cross section of

shear deformable curved beam subjected to the external

forces can be evaluated using Hooke’s law.

4. Isoparametric Curved Beam Element

For comparison, an isoparametric thin-walled curved

beam based on Eqs. (15) and (16) is presented in this

Section. This element has two nodes per element and

seven nodal degrees of freedom per node (see Fig. 3).

The coordinate and all the displacement parameters of the

beam element can be linearly interpolated with respect to

the nodal coordinates and displacements, respectively.

Substituting the shape functions, cross-sectional properties

into Eqs. (15) and (16) and integrating along the element

length, and then the total potential energy of thin-walled

curved beam element is obtained in matrix form as

F1

αF2

αF3

αM

1

αM

2

αM

3

αMφ

α

KSX 0( )E 1––

SX l( )E 1–=

e11

d2x2d6

–1

R---d

7x3d10

x2

R----d

11φd

14+ + + +⎝ ⎠

⎛ ⎞ RR x

3+

------------=

Figure 3. Nodal displacement vector of a thin-walledcurved beam element.

Figure 4. Nodal force vector of a thin-walled curvedbeam element.

(a) Geometry of a curved beam (b) x3-monosymmetric cross section

A = 10 cm2, E = 73000 N/cm2, G = 28000 N/cm2, J = 0.833333 cm4, e2 = 0 cm, e3 = −6.89231 cm I2 = 68.26667 cm4,

I3 = 34.66667 cm4, I222 = 40.96 cm

5, I333 = 17.06667 cm5, Iφ = 1856.85333 cm

6 Iφ3 = 238.93333 cm5, Iφ23= 327.68 cm

6, Iφφ2= 3833.856 cm

7, = 0.93577 cm2, = 6.82667 cm2, = 27.80720 cm4, l = 100 cm

(c) Material and section properties

A2

s

A3

s

Ar

s

Figure 5. Cantilevered beam with x3-monosymmetric cross section.


(41)

where Ke and Kg = the element’s elastic and geometric

stiffness matrices in local coordinates, respectively.

Stiffness matrices are evaluated using a reduced Gauss

numerical integration scheme. Here, it should be noticed

that the element displacement and force vectors of an

isoparametric curved beam are identical to those of an

exact curved beam but the interpolation functions are

different.

5. Numerical Examples

To illustrate the accuracy and the practical usefulness

of this study, numerical solutions for the elastic analysis

of shear deformable thin-walled symmetric and non-

symmetric curved beams are presented and compared

with the results by available references as well as by

SAP2000’s shell elements (SAP 2000 1995).

5.1. Cantilevered curved beam with x3-monosymmetric

section

The thin-walled cantilevered curved beam with x3-

monosymmetric section subjected to a vertical force 50 N

at the free end and its material and sectional properties

are shown in Fig. 5. The length of a curved beam is

100 cm. The horizontal, vertical displacements and angle

of rotation at the free end of the curved beam are

evaluated and presented in Table 1. For comparison, the

results by finite element solutions using isoparametric

curved beam elements and the analytical solution

obtained from Castigliano’s energy theorem (Tauchert

1974) considering shear deformation effect are together

presented. The exact responses at the free end of

cantilevered curved beam using the energy theorem

subjected to a vertical force P are given as

ΠT

1

2---U

e

TK

eK

g+( )U

eU

e

TFe

–=

Figure 6. Convergence of normalized displacements at thefree end of the cantilevered beam under a vertical force.

Table 1. Horizontal, vertical displacements and angle of rotation at the free end of the x3-monosymmetric cantileveredbeam under a vertical force (cm, rad.×10−2)

This studyTauchert(1974)Exact curved

beam element

Isoparametric curved beam element

2 6 10 20 40

Ux

-1.315 -1.129 -1.292 -1.307 -1.313 -1.314 -1.313

Uz 2.066 1.694 2.019 2.049 2.061 2.065 2.069

ω2 -4.105 -3.801 -4.069 -4.092 -4.102 -4.104 -4.105

(a) Non-symmetric L-shaped cross section

A = 24.1935 cm2, E = 20684.28 kN/cm2, G = 7955.49 kN/cm2, J = 13.00723 cm4, e2 = −4.23333 cm e3 = 1.05833 cm, I2 = 81.29520 cm

4, I3 = 433.57440 cm4, I23 = 108.39360 cm

4, I222 =-229.43312 cm5 I223 = −229.43312 cm

5,Iφ = 2913.80064 cm

6, Iφ2 = −453.86624 cm5, Iφ3 = −917.73248 cm

5 Iφ22 =1214.08360 cm6, Iφ23 = 971.26688 cm

6, Iφφ2 =-6167.54468 cm

7, = 12.94278 cm2 = 5.96526 cm2, = 0.0 cm4, R = 914.4 cm, l = 609.6 cm

(b) Material and section properties

Figure 7. L-shaped section of the clamped curved girder.

A2

sA3

sA

r

s


(42a)

(42b)

(42c)

The present results are found to be in excellent

agreement with finite element results using a large

number of curved beam elements and analytical solution

(42). Convergence behaviors of the normalized horizontal

and vertical displacements at the free end of the

cantilevered curved beam, as the number of finite elements

increase, are displayed in Fig. 6. As can be seen in Fig.

6, the present solutions using only single element

coincide well with the solutions using 16 curved beam

elements.

5.2. Fixed L-shaped curved girder

In this example, we consider the curved girder with

non-symmetric L-shaped section as shown in Fig. 7. The

girder is fixed at two ends and subjected to an out-of-

plane lateral force 4.45 kN (1000 lb) at the midspan. As

the boundary condition is fixed at two ends, 2 curved

beam elements using exact static stiffness matrix are

used. Fig. 8 shows the lateral displacement at the corner

of the L-shaped cross section along the curved girder. For

comparison, by considering the symmetry, the results

obtained from 20 isoparametric curved beam elements

and 8 HMC2 curved beam elements by Gendy and Saleeb

(1992), based on mixed variational formulation and 24

quadrilateral shell elements developed by Saleeb et al.

(1990), are presented. From Fig. 8, it can be found that

present solutions are in good agreement with those

obtained from isoparametric and HMC2 curved beam

elements and shell elements.

5.3. Cantilevered curved beam with doubly symmetric

section

The purpose of this example is to show the practical

usefulness of the proposed numerical method by

calculating the normal stresses at the arbitrary point of

thin-walled cross section. Fig. 9 shows the cantilevered

curved beam with doubly symmetric cross section

Ux

PR3

2EÎ2

-----------–PR

2EA----------

PR

2GA3

-------------–+=

Uz

πPR3

4EÎ2

------------–πPR

4EA----------

πPR

4GA3

-------------–+=

ω2

PR2

EÎ2

---------=

Figure 8. Lateral displacement at the corner of clampedL-shaped girder.

A = 8 cm2, E = 73000 N/cm2, G = 28000 N/cm2, J = 0.66667 cm4, I2 = 116.66667 cm4, I3 = 2.25 cm

4 Iφ = 56.25 cm6,

Iφ23 = −56.25 cm6, = 2.5 cm2, = 4.77932 cm2, = 62.5 cm4, l = 100 cm

(c) Material and section properties

Figure 9. Cantilevered curved beam with a doubly symmetric cross section.

A2

s

A3

s

Ar

s

(a) Geometry of a curved beam (b) Doubly symmetric cross section


subjected to a vertical force 100N at the free end with its

material and sectional properties. Table 2 gives the

horizontal, vertical displacements and angle of rotation at

the centroid of free end of curved beam. For comparison,

the results by curved beam elements and 1800 SAP2000’s

shell results are given together. From Table 2, the excellent

agreement between these results is observed.

Next, the normal stresses at the points and of the mid-

Table 2. Horizontal, vertical displacements and angle of rotation at the free end of the doubly symmetric cantileveredbeam under a vertical force (cm, rad.×10-2)

This study1800 shellelementsExact curved

beam element


2 6 10 20 40

Ux -1.539 -1.323 -1.512 -1.529 -1.536 -1.538 -1.543

Uz 2.417 1.984 2.363 2.397 2.412 2.415 2.421

ω2 -4.759 -4.407 -4.717 -4.743 -4.755 -4.758 -4.994

Table 3. Normal stresses at the mid-section of the doubly symmetric cantilevered beam under a vertical force (N/cm2)

Point


beam element


2 6 10 20 40

① -178.88 -308.51 -198.28 -186.04 -180.69 -179.33 -177.09

② 209.37 -11.978 177.77 197.74 206.43 208.63 207.20

(a) Non-symmetric cross section

A = 9.6 cm2, E = 73000 N/cm2, G = 28000 N/cm2, J = 1.152 cm4, I2 = 102.4 cm4, I3 = 25.6 cm

4 I23 = 38.4 cm4, Iφ =

256.0 cm6, Iφ22 =-204.8 cm6, Iφ23 = -256.0 cm

6, Iφ33 =-204.8 cm6 = 3.65991 cm2, = 4.56380 cm2, =

47.05882 cm4, l = 100 cm


Figure 10. Non-symmetric cross section of the cantilevered beam under a vertical force.

A2

s

A3

s

Ar

s

Table 4. Displacements at the free end of the non-symmetric cantilevered beam under a vertical force (cm, rad.×10-2)


beam element


2 6 10 20 40

Ux -3.971 -3.409 -3.901 -3.945 -3.964 -3.969 -3.928

Uy -9.153 -7.441 -8.939 -9.075 -9.134 -9.148 -9.172

Uz 6.237 5.115 6.097 6.186 6.225 6.234 6.168

ω1 15.45 12.73 15.11 15.33 15.42 15.45 14.88

ω2 -12.39 -11.48 -12.28 -12.35 -12.38 -12.39 -12.47

ω3 -8.756 -8.734 -8.758 -8.757 -8.756 -8.756 -9.013


section of curved beam (Fig. 9b) are evaluated and

compared with the results by curved beam elements and

SAP2000’s shell elements in Table 3. In evaluating the

normal stresses by isoparametric curved beam element in

this and the next Sections, we take the average value of

normal stresses for two elements. From Table 3, it is

observed that the present solutions using only one

element are in good agreement with those by 40 curved

beam elements and shell elements. It may be remarked

particularly from Tables 2 and 3 that at least 6 isoparametric

curved beam elements are necessary in calculating

displacements. Still, 20 isoparametric beam elements are

demanded for the reasonably good results in calculating

normal stresses.

5.4. Cantilevered curved beam with non-symmetric

section

The spatially coupled cantilevered curved beam with

non-symmetric cross section subjected to a vertical force

100 N at the centroid of free end is considered as shown

in Figs. 9a and 10a. The length is 100 cm and the material

and sectional properties are given in Fig. 10. Tables 4 and

Table 5. Normal stresses at the mid-section of the non-symmetric cantilevered beam under a vertical force (N/cm2)

Point


beam element


2 6 10 20 40

① 186.06 -303.20 116.82 161.52 181.01 185.94 183.15

② -374.59 -732.76 -430.41 -397.14 -382.59 -378.90 -376.14

③ 426.23 -115.81 351.49 400.80 422.27 427.70 429.43

④ -214.70 -607.31 -273.86 -237.57 -221.72 -217.70 -219.06

(a) x2-monosymmetric cross section

A = 9 cm2, E = 73000 N/cm2, G = 28000 N/cm2, J = 0.75 cm4, e2 = 2.30065 cm, e3 = 0 cm I2 = 141.66667 cm4,

I3 = 14.22222 cm4, I233 = −74.07407 cm

5, Iφ = 1000.82305 cm6 Iφ2 = −325.92593 cm

5, Iφ23 = −421.39918 cm6,

Iφφ3 = −1816.55235 cm7, = 2.21071 cm2 = 4.34717 cm2, = 74.10624 cm4, l = 100 cm


Figure 11. x2-monosymmetric cross section of the cantilevered beam.

A2

s

A3

s

Ar

s

(a) Cantilevered curved beam (b) Fixed curved beam

Figure 12. Cantilevered and fixed curved beams subjected to constant initial axial force and external force.


5 give comparisons of solutions using one exact curved

beam element with results using isoparametric curved beam

elements and 1440 shell elements of SAP2000. The

results using one element coincide well with the solutions

by isoparametric beam elements as the number of

elements increase and also are in good agreement with

those by shell elements. The maximum difference

between the exact solutions and the results by shell

elements is 1.99% at point . Also, it can be observed from

Table 5 that at least 40 isoparametric beam elements are

demanded for accurate calculation of normal stresses in

the case of the non-symmetric curved beam.

5.5. Cantilevered and fixed curved beams with x2-

monosymmetric section considering initial forces

In our final example, the spatially coupled elastic

Table 6. Displacements at the free end of the x2-monosymmetric cantilevered beam under axial and vertical forces (cm,rad.×10-2)

This study

Exact curved beam element


2 6 10 20 40

Ux

[-0.9048]-0.9670(-1.073)

[-0.8074]-0.8584

(-0.9338)

[-0.8930]-0.9537(-1.054)

[-0.9005]-0.9622(-1.066)

[-0.9037]-0.9658(-1.071)

[-0.9045]-0.9667(-1.072)

Uy

[-3.012]-3.939

(-5.566)

[-3.934]-4.778

(-6.048)

[-3.138]-4.052

(-5.615)

[-3.058]-3.980

(-5.583)

[-3.023]-3.949

(-5.570)

[-3.015]-3.941

(-5.567)

Uz

[1.455]1.519

(1.625)

[1.219]1.285

(1.383)

[1.425]1.490

(1.594)

[1.444]1.508

(1.614)

[1.452]1.516

(1.622)

[1.454]1.518

(1.624)

ω1

[-20.10]-22.51

(-26.59)

[-17.58]-20.14

(-23.96)

[-19.79]-22.22

(-26.24)

[-19.98]-22.41

(-26.46)

[-20.07]-22.49

(-26.56)

[-20.09]-22.51

(-26.58)

ω2

[-2.916]-2.995

(-3.117)

[-2.753]-2.849

(-2.986)

[-2.896]-2.977

(-3.101)

[-2.909]-2.988

(-3.111)

[-2.914]-2.993

(-3.116)

[-2.915]-2.994

(-3.117)

ω3

[-11.72]-14.37

(-18.96)

[-11.33]-13.51

(-16.78)

[-11.69]-14.27

(-18.64)

[-11.71]-14.33

(-18.84)

[-11.72]-14.36

(-18.93)

[-11.72]-14.37

(-18.95)

f[0.4211]0.4498

(0.4966)

[0.4128]0.4494

(0.5033)

[0.4200]0.4496

(0.4970)

[0.4207]0.4497

(0.4967)

[0.4210]0.4498

(0.4966)

[0.4211]0.4498

(0.4966)

[ ], the results with an axial tensile force 10 N.( ), the results with an axial compressive force 10 N.

Table 7. Lateral, vertical displacements and twisting angle at the mid-section of the x2-monosymmetric fixed beam underaxial and vertical forces (cm, rad.×10-2)

This study

Exact curved beam element


6 10 20 40 60

Uy

[-2.406]-2.999

(-3.879)

[-1.666]-1.969

(-2.376)

[-2.125]-2.595

(-3.266)

[-2.334]-2.894

(-3.718)

[-2.388]-2.973

(-3.838)

[-2.398]-2.987

(-3.861)

Uz

[-3.586]-3.697

(-3.833)

[-3.045]-3.101

(-3.164)

[-3.386]-3.474

(-3.578)

[-3.535]-3.641

(-3.767)

[-3.573]-3.683

(-3.816)

[-3.580]-3.691

(-3.826)

ω1

[0.7310]0.7952

(0.8801)

[0.6034]0.6383

(0.6803)

[0.6842]0.7362

(0.8024)

[0.7192]0.78020.8600

[0.7281]0.7914

(0.8750)

[0.7297]0.7935

(0.8778)

[ ], the results with an axial tensile force 500 N.( ), the results with an axial compressive force 500 N.


analysis of cantilevered and fixed curved beams is

performed in which the constant initial axial forces act

along the centroid of beam. Fig. 11 shows the x2-

monosymmetric cross section and its geometric and

material data. It is noted that in-plane and out-of-plane

behaviors of this curved beam are coupled because the

cross section is monosymmetric for x2 axis. Figs. 12(a)

and 12(b) show the configuration of cantilevered and

fixed curved beams, respectively, where the subtended

angle is taken to be 90o and length of beams is 100 cm for

both curved beams. First, we consider a cantilevered

curved beam subjected to a vertical force 50 N at the

centroid of free end with 10 N as the intial axial force.

Next, a fixed curved beam subjected to a vertical force

-1000 N at the centroid of free end is considered with

500 N as the intial axial force. Tables 6 and 7 give

comparison of solutions using isoparametric curved beam

elements with the results using only one exact element in

case of a cantilevered beam and two exact elements in

case of a fixed beam, respectively. The excellent agreement

is observed between results using the exact element and

isoparametric elements.

6. Conclusions

An exact curved beam element for the spatially coupled

static analysis of shear deformable thin-walled curved

beam subjected to initial axial force is presented for the

first time. For this, the equilibrium equations are transformed

into first-order simultaneous ordinary differential equations

by introducing 14 displacement parameters and a generalized

linear eigenproblem having complex eigenvalues is

considered. Then, displacement functions of displacement

parameters are exactly evaluated and finally, the exact

element stiffness matrix is determined using the force-

deformation relations. In case of an isoparametric beam

element, linear interpolation functions are adopted for

displacement parameters.

Through the numerical examples, it turns out that the

displacements, the twisting and rotational angles, and the

normal stresses at the arbitrary point of non-symmetric

cross section for the spatially coupled curved beam by

this study are in good agreement with the finite element

solutions as well as those obtained from the shell

elements. Particularly, the exact curved beam element

gives the accurate stress at the arbitrary point along the

length though only one element is used.

Acknowledgments

This work is a part of a research project supported by

Korea Ministry of Construction & Transportation through

Korea Bridge Design & Engineering Research Center at

Seoul National University. The paper was finalized while

the first author visits and conducts research at the

University of Maryland, College Park, USA. The authors

express their gratitude for the financial support.

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Appendix I.

1) Components of matrix A

where

2) Components of matrix B

where

k1

k2 k3 k4 k5 k6 k7 k8

k1

k9 k10 k11

k1

k12 -k10 k13 k14 K15 k16

A =k1

k17 -k11 k18 k16 -k4 k19

K1

k6 k10 k20 k21 k22 -k16 k23

k1

k4 k24 k25 k26 k20 k27 k28

k1

k8 k11 k28 -k16 k23 k29 k30

k1

1.0 k2

EA–EÎ2

R2

------- k3

GA23

R------------– , k

4

EÎ23

R---------= , k

5

GÂ3

R----------= , k

6

EÎ2

R-------= , k

7

GA3r

R-----------= , k

8

EÎφ2

R----------= k

9GA

2F

o

1+ k

10GA

23=,=,=,–=,=

k11

GA2r

k12

1

R--- EA

EÎ2

R2

------- GA3

Fo

1+ + +⎝ ⎠

⎛ ⎞k13

EÎ23

R2

--------- k14

GA3

Fo

1+( )– k

15

EÎ2

R2

------- k16

GA3r

R-----------– k

17

1

R---–EÎ23

R--------- GA

3r–⎝ ⎠

⎛ ⎞=,=,=,=,=,=,=

k18

1

R---EÎ3

R------- GA

r+⎝ ⎠

⎛ ⎞– k19

GJ– GAr

– β Fo1

– k20

EÎ23 k21 GA3 k22 EÎ2– k23 EÎφ2– k24 GA2–GA

2r

R-----------+=,=,=,=,=,=,=

k25

EÎ3– k26 GA23–GA

3r

R-----------+ k

27

GJ

R-------

GAr

R--------- GA

2r–

βR--- F

o

1+ + k

28EÎφ3 k29 GAr k30 EÎφ–=,=,=,=,=,=

b1

b2 b3 b4 b5 b6 b7

b1

b8 b9 b10 b11

b12 b13 b14 b15 b16

b6 b7 b8 b9 b10 b11

B =b1

b17 b15 b18 b19 b20

b1

b5 -b12 b21 -b14 b22 -b18

b1

b3 b23 -b6 -b12 -b22 b24

b1

b7 b25 b26 -b18 b27 b28

(A-1)

(A-2)


3) Components of matrix S

Where

b1

1.0 b2

1

R---– GA

3F

o

1+( ) b

3

GA23

R------------

GA3r

R-----------+– b,

4

1

R--- EA

EÎ2

R2

------- Fo

1+ +⎝ ⎠

⎛ ⎞= b,5

GA3

R----------= b,

6

EÎ23

R2

---------= b,7

GA3r

R-----------= b,

8

GA23

R------------==,=,=

b9

GA2

GA2r

R-----------– b

10GA

23b11

,–= GA2r

– b,=12

GA23

–GA

3r

R-----------+ b

13

1

R2

-----– EAEÎ2

R2

-------+⎝ ⎠⎛ ⎞

b14

GA3

b15

EÎ23

R2

---------=,=,=,=,=

b16

GA3r

EÎφ2

R2

----------+ b17

GJ

R------- GA

2r–

βR--- F

o

– b18

GA3r

b19

EÎ2

R2

-------– b20

EÎφ3

R---------- GA

r+– b

21

EÎ2

R2

-------– b22

EÎ23

R---------=,=,=,=,=,=,=

b23

GJ

R------- GA

2–

GAr

R2

---------–2GA

2r

R---------------

β

R2

----- Fo

1–+– b

24

GAr

R---------– GA

2r+ b

25GA

2r

GAr

R---------– k

26

EÎφ2

R2

---------- b27

EÎφ3

R---------- b

28GA

r–=,=,=,=,=,=

s1 s2 s3 s4 s5 s6

s7 s8 s9 s10 s10 s11 s11

s12 s10 s13 s14 s15 s16 s16

S = s17 s11 s18 s16 s16 s19 s20s4 s21 s22 s23 s2 s24

s2 s25 -s5 s21 s26 s27

s6 s27 s28 s24 s29 s30

s1

EAEÎ2

R2

-------+= s2

EÎ23

R---------= s

3

1

R--- EA

EÎ2

R2

-------+⎝ ⎠⎛ ⎞= s

4

EÎ2

R-------–= s

5

EÎ23

R2

---------= s6

EÎφ2

R---------- s

7

GA23

R------------– s

8GA

2F

o

1+ s

9GA

2–

GA2r

R-----------+=,=,=,=, , , , ,

s10

GA23

= s11

GA2r

= s12

1

R---– GA

3F

o

1+( )= s

13GA

23–

GA2r

R-----------+= s

14GA

3F

o

1+= s

15GA

3, s

16GA

3r= s

17

GA3r

R-----------–=,=, , , , ,

s18

GJ

R-------

GAr

R--------- GA

2r–

βR--- F

o

1+ += s

19GJ GA

rβ Fo

1+ += s

20GA

r= s

21EÎ23–= s22

EÎ2

R2

-------–= s23

EÎ2, s24 EÎφ2= s25 EÎ3=,=, , , , ,

s26

EÎ3

R-------–= s

27EÎφ3–= s28

EÎφ2

R2

----------–= s29

EÎφ3

R----------= s

30EÎφ=, , , ,

(A-3)

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